Try `with=FALSE` :

    dt[,-1,with=FALSE]

---

As an aside, feature request #416 is related :

[Add *not join* `DT[-J(...)]`, and *not columns* `DT[,-&quot;colC&quot;,with=FALSE]`.][1] 


  [1]: https://github.com/Rdatatable/data.table/issues/416

The previous answer works, but here is the answer with the updated methods:

    ## CONVERT TO ZOO
    
    #### old method
    zooObj = zoo(x=dt[,-1,with=FALSE], order.by=dt$mydates)
    
    #### alternatives
    zooObj = zoo(x=dt[,!&#39;mydates&#39;], order.by=dt$mydates)
    zooObj = zoo(x=dt[,-&#39;mydates&#39;], order.by=dt$mydates)
    
    #### and if column name is a variable
    myvar = &#39;mydates&#39;
    zooObj = zoo(x=dt[,!..myvar], order.by=dt[[myvar]])
    zooObj = zoo(x=dt[,-..myvar], order.by=dt[[myvar]])

As an extension, if you wanted to remove more than one column from the `data.table`, try one of the following:

    dt[,-c(&#39;value1&#39;, ..myvar)]
    dt[, .SD, .SDcols = !c(&#39;value1&#39;, (myvar))]

Hey all i am trying to figure out how to go about inserting a new record using the following query:

    SELECT user.id, user.name, user.username, user.email, 
      IF(user.opted_in = 0, &#39;NO&#39;, &#39;YES&#39;) AS optedIn  
    FROM 
      user
      LEFT JOIN user_permission AS userPerm ON user.id = userPerm.user_id
    ORDER BY user.id;

My `INSERT` query so far is this:

    INSERT INTO user 
    SELECT * 
    FROM user 
      LEFT JOIN user_permission AS userPerm ON user.id = userPerm.user_id;

However, i  am not sure how to do `VALUE(&#39;&#39;,&#39;&#39;,&#39;&#39;,&#39;&#39;, etc etc)` when using left and inner joins.

So what i am looking to do is this:

`User` table:

    id    | name       | username    | password                 | OptIn
    --------------------------------------------------------------------
    562     Bob Barker   bBarker       BBarker@priceisright.com   1

And also the `user_permission` table

    user_id   | Permission_id
    -------------------------
    562         4

**UPDATE**
So like this?

    INSERT INTO user (name, username, password, email, opted_in) VALUES (&#39;Bbarker&#39;,&#39;Bbarker&#39;,&#39;blahblahblah&#39;,&#39;Bbarker@priceisright.com&#39;,0);
    INSERT INTO user_permission (user_id, permission_id) VALUES (LAST_INSERT_ID(),4);

Insert using LEFT JOIN and INNER JOIN

I have a div element wrapping other div elements like so:

    &lt;div style=&quot;overflow:hidden&quot;&gt;
        &lt;div id=&quot;a&quot;&gt;&lt;/div&gt;
        &lt;div id=&quot;b&quot;&gt;&lt;/div&gt;
    &lt;/div&gt;

I have other css rules that manage the dimensions of the outer div.  In my actual code,  I want to position the div#a exactly 10 px below the outer div.  However, I want div#b to still be cut off by the outer div&#39;s overflow:hidden.

What is the best way to achieve this?


How to ignore parent element&#39;s overflow:hidden in css

When using `data.table` is it possible to return all the columns except one, like in `data.frame`?

If the answer is no, does anyone have an elegant way to transform a multiple time series `data.table` to a `zoo` or other time series object?

Consider the following example:

    library(data.table)
    library(zoo)
    
    ## DEFINE DATA
    set.seed(1)
    dt = data.table(
        mydates = as.Date(&quot;2012-01-01&quot;) + 1:9, 
    	value1 = sort(rpois(9, 6)),
    	value2 = sort(rpois(9, 6)),
    	value3 = sort(rpois(9, 6)),
    	value4 = sort(rpois(9, 6)),
    	value5 = sort(rpois(9, 6)))
    
    ## CONVERT TO DATA FRAME
    df = as.data.frame(dt)
    
    ## CONVERT TO ZOO
    zooObj = zoo(df[,-1], df$mydates)

    ## EXAMPLE OF DESIRED RESULTS
    plot(zooObj, col=1:ncol(zooObj))

How would I do that without `df = as.data.frame(dt)`?


How to exclude one column from data.table OR convert to data.table to MTS

When using <code>data.table</code> is it possible to return all the columns except one, like in <code>data.frame</code>?
If the answer is no, does anyone have an elegant way to transform a multiple time series <code>data.table</code> to a <code>zoo</code> or other time series object?
Consider the following example:
<pre><code class="hljs language-ini">library(data.table)
library(zoo)

## DEFINE DATA
set.seed(1)
dt = data.table(
 mydates = as.Date("2012-01-01") + 1:9, 
	value1 = sort(rpois(9, 6)),
	value2 = sort(rpois(9, 6)),
	value3 = sort(rpois(9, 6)),
	value4 = sort(rpois(9, 6)),
	value5 = sort(rpois(9, 6)))

## CONVERT TO DATA FRAME
df = as.data.frame(dt)

## CONVERT TO ZOO
zooObj = zoo(df[,-1], df$mydates)

## EXAMPLE OF DESIRED RESULTS
plot(zooObj, col=1:ncol(zooObj))
</code></pre>
How would I do that without <code>df = as.data.frame(dt)</code>?

Sourcing files using a relative path is useful when dealing with large codebases. Other programming languages have well-defined mechanisms for sourcing files using a path relative to the directory of the file being sourced into. An example is Ruby&#39;s [`require_relative`](http://extensions.rubyforge.org/rdoc/classes/Kernel.html). What is a good way to implement relative path sourcing in R?

Below is what I pieced together a while back using various recipes and R forum posts. It&#39;s worked well for me for straight development but is not robust. For example, it breaks when the files are loaded via the `testthat` library, specifically `auto_test()`. `rscript_stack()` returns `character(0)`.

	# Returns the stack of RScript files
	rscript_stack &lt;- function() {
	  Filter(Negate(is.null), lapply(sys.frames(), function(x) x$ofile))
	}

	# Returns the current RScript file path
	rscript_current &lt;- function() {
	  stack &lt;- rscript_stack()
	  r &lt;- as.character(stack[length(stack)])
	  first_char &lt;- substring(r, 1, 1)
	  if (first_char != &#39;~&#39; &amp;&amp; first_char != .Platform$file.sep) {
	    r &lt;- file.path(getwd(), r)
	  }
	  r
	}

	# Sources relative to the current script
	source_relative &lt;- function(relative_path, ...) {
	  source(file.path(dirname(rscript_current()), relative_path), ...)
	}

Do you know of a better `source_relative` implementation?

R: sourcing files using a relative path

I have the following data frame

    x &lt;- read.table(text = &quot;  id1 id2 val1 val2
    1   a   x    1    9
    2   a   x    2    4
    3   a   y    3    5
    4   a   y    4    9
    5   b   x    1    7
    6   b   y    4    4
    7   b   x    3    9
    8   b   y    2    8&quot;, header = TRUE)

I want to calculate the mean of val1 and val2 grouped by id1 and id2, and simultaneously count the number of rows for each id1-id2 combination. I can perform each calculation separately: 

    # calculate mean
    aggregate(. ~ id1 + id2, data = x, FUN = mean)

    # count rows
    aggregate(. ~ id1 + id2, data = x, FUN = length)

In order to do both calculations in one call, I tried

    do.call(&quot;rbind&quot;, aggregate(. ~ id1 + id2, data = x, FUN = function(x) data.frame(m = mean(x), n = length(x))))

However, I get a garbled output along with a warning:

    #     m   n
    # id1 1   2
    # id2 1   1
    #     1.5 2
    #     2   2
    #     3.5 2
    #     3   2
    #     6.5 2
    #     8   2
    #     7   2
    #     6   2
    # Warning message:
    #   In rbind(id1 = c(1L, 2L, 1L, 2L), id2 = c(1L, 1L, 2L, 2L), val1 = list( :
    #   number of columns of result is not a multiple of vector length (arg 1)

I could use the plyr package, but my data set is quite large and plyr is very slow (almost unusable) when the size of the dataset grows.

How can I use `aggregate` or other functions to perform several calculations in one call?
 





Apply several summary functions on several variables by group in one call

I&#39;m struggling a great deal to modify the legend in my plot. Here is a reproducible example:

    dtt &lt;- structure(list(model = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 3L, 3L, 3L, 3L, 3L, 3L), .Label = c(&quot;ma&quot;, &quot;mb&quot;, &quot;mc&quot;), class = &quot;factor&quot;), year = c(2005L, 2006L, 2007L, 2008L, 2009L, 2010L, 2005L, 2006L, 2007L, 2008L, 2009L, 2010L, 2005L, 2006L, 2007L, 2008L, 2009L, 2010L), V = c(0.16, 0.14, 0.11, 0.13, 0.15, 0.16, 0.24, 0.17, 0.12, 0.13, 0.15, 0.15, 0.2, 0.16, 0.11, 0.12, 0.12, 0.15), lower = c(0.11, 0.11, 0.07, 0.09, 0.11, 0.12, 0.16, 0.12, 0.04, 0.09, 0.09, 0.11, 0.14, 0.1, 0.07, 0.08, 0.05, 0.1), upper = c(0.21, 0.19, 0.17, 0.17, 0.19, 0.2, 0.29, 0.23, 0.16, 0.17, 0.16, 0.2, 0.26, 0.27, 0.15, 0.16, 0.15, 0.19)), .Names = c(&quot;model&quot;, &quot;year&quot;, &quot;V&quot;, &quot;lower&quot;, &quot;upper&quot;), class = &quot;data.frame&quot;, row.names = c(NA, -18L))

My plot is generated like this:

    ggplot(dtt, aes(x=year, y=V, group = model, colour = model, ymin = lower, ymax = upper)) +
        geom_ribbon(alpha = 0.35, linetype=0)+ 
    	geom_line(aes(linetype=model), size = 1.5) +       
    	geom_point(aes(shape=model), fill = &quot;white&quot;, size = 4)  +      
        theme(legend.position=c(.6,0.8)) +
        theme(legend.background = element_rect(colour = &#39;black&#39;, fill = &#39;grey90&#39;, size = 1, linetype=&#39;solid&#39;))

which produces this:
![enter image description here][1]

Now, what I would like to do is

1. change the title of the legend
2. change the order in which the legend items appear
3. change the text of the legend items.

I have fiddled around for hours trying to do this, but without much success. The best I have managed so far is to add this:

    scale_colour_hue(name = &quot;Model 1&quot;,
        breaks=c(&quot;mb&quot;, &quot;ma&quot;, &quot;mc&quot;),
        labels=c(&quot;MBB&quot;, &quot;MAA&quot;, &quot;MCC&quot;))

But it produces this abomination:
![enter image description here][2]

As you see, there is now an extra unneeded legend, and the shapes in the legend do not match those in the plot !

Finally, I would like to graphics in the legend to indicate that the blue and green lines are dashed, not solid - but I have no idea at all how to do that.

Any assistance would be highly appreciated,

  [1]: http://i.stack.imgur.com/xTgJi.png
  [2]: http://i.stack.imgur.com/je9Bi.png

ggplot legends - change labels, order and title

I&#39;m trying to use knitr to generate a report that performs the same set of analyses on different subsets of a data set. The project contains two Rmd files: the first file is a master document that sets up the workspace and the document, the second file only contains chunks that perform the analyses and generates associated figures.

What I would like to do is knit the master file, which would then call the second file for each data subset and include the results in a single document. Below is a simple example.


Master document:

    # My report

    ```{r}
    library(iterators)
    data(mtcars)
    ```

    ```{r create-iterator}
    cyl.i &lt;- iter(unique(mtcars$cyl))
    ```

    ## Generate report for each level of cylinder variable
    ```{r cyl4-report, child=&#39;analysis-template.Rmd&#39;}
    ```

    ```{r cyl6-report, child=&#39;analysis-template.Rmd&#39;}
    ```

    ```{r cyl8-report, child=&#39;analysis-template.Rmd&#39;}
    ```

analysis-template.Rmd:

    ```{r, results=&#39;asis&#39;}
    cur.cyl &lt;- nextElem(cyl.i)
    cat(&quot;###&quot;, cur.cyl)
    ```

    ```{r mpg-histogram}
    hist(mtcars$mpg[mtcars$cyl == cur.cyl], main = paste(cur.cyl, &quot;cylinders&quot;))
    ```

    ```{r weight-histogam}
    hist(mtcars$wt[mtcars$cyl == cur.cyl], main = paste(cur.cyl, &quot;cylinders&quot;))
    ```

The problem is knitr does not allow for non-unique chunk labels, so knitting fails when `analysis-template.Rmd` is called the second time. This problem could be avoided by leaving the chunks unnamed since unique labels would then be automatically generated. This isn&#39;t ideal, however, because I&#39;d like to use the chunk labels to create informative filenames for the exported plots.

----------

A potential solution would be using a simple function that appends the current cylinder to the chunk label:

    ```r{paste(&#39;cur-label&#39;, cyl, sep = &quot;-&quot;)}
    ```
But it doesn&#39;t appear that knitr will evaluate an expression in the chunk label position.

----------

I also tried using a custom [chunk hook](http://yihui.name/knitr/hooks#chunk_hooks) that modified the current chunk&#39;s label: 

    knit_hooks$set(cyl.suffix = function(before, options, envir) {
        if (before) options$label &lt;- &quot;new-label&quot;
    })

But changing the chunk label didn&#39;t affect the filenames for generated plots, so I didn&#39;t think knitr was utilizing the new label.


----------

Any ideas on how to change chunk labels so the same child document can be called multiple times? Or perhaps an alternative strategy to accomplish this? 

R knitr: Possible to programmatically modify chunk labels?

I need to remove the last number in a groups of vectors, i.e.:

    v &lt;- 1:3
    v1 &lt;- 4:8

should become:

    v &lt;- 1:2
    v1 &lt;- 4:7


remove the last element of a vector

The data.table has a nice feature that suppresses output to the head and tail of the table.

Is it possible to view / print more than 100 rows at once?

    library(data.table)
    ## Convert the ubiquitous &quot;iris&quot; data to a data.table
    dtIris = as.data.table(iris)
    ## Printing 100 rows is possible
    dtIris[1:100, ]
    ## Printing 101 rows is truncated
    dtIris[1:101, ]

I often have data.table results that are somewhat large (e.g. 200 rows) that I just want to view.


Possible to print more than 100 rows of a data.table?

In R, I have an operation which creates some `Inf` values when I transform a dataframe.  

I would like to turn these `Inf` values into `NA` values.  The code I have is slow for large data, is there a faster way of doing this? 

Say I have the following dataframe: 

    dat &lt;- data.frame(a=c(1, Inf), b=c(Inf, 3), d=c(&quot;a&quot;,&quot;b&quot;))

The following works in a single case: 

     dat[,1][is.infinite(dat[,1])] = NA

So I generalized it with following loop

    cf_DFinf2NA &lt;- function(x)
    {
    	for (i in 1:ncol(x)){
    		  x[,i][is.infinite(x[,i])] = NA
    	}
    	return(x)
    }

But I don&#39;t think that this is really using the power of R. 

Cleaning `Inf` values from an R dataframe

I&#39;d like to reorder columns in my `data.table` `x`, given a character vector of column names, `neworder`:

    library(data.table)
    x &lt;- data.table(a = 1:3, b = 3:1, c = runif(3))
    neworder &lt;- c(&quot;c&quot;, &quot;b&quot;, &quot;a&quot;)

Obviously I could do:

    x[ , neworder, with = FALSE]
    # or
    x[ , ..neworder]
    #            c b a
    # 1: 0.8476623 3 1
    # 2: 0.4787768 2 2
    # 3: 0.3570803 1 3

but that would require copying the entire dataset again. Is there another way to do this?




How to reorder data.table columns (without copying)

Let&#39;s say I have the following ``data.table`` in ``R``:

      library(data.table)
      DT = data.table(x=rep(c(&quot;b&quot;,&quot;a&quot;,&quot;c&quot;),each=3), y=c(1,3,6), v=1:9)

I want to order it by two columns (say columns ``x`` and ``v``). I used this:

     DT[order(x,v)] # sorts first by x then by v (both in ascending order)

But now, I want to sort it by ``x`` (in decreasing order) and have the following code:

      DT[order(-x)] #Error in -x : invalid argument to unary operator

Therefore, I think this error is due to the fact that ``class(DT$x)=character``. Could you give me any suggestion in order to solve this issue?

I know I can use ``DT[order(x,decreasing=TRUE)]``, but I want to know the syntax to sort by several columns using both ways (some decreasing, some increasing) at the same time. 

Note that if you use ``DT[order(-y,v)]`` the result is ok, but if you use ``DT[order(-x,v)]`` there is an error. So, my question is: how to solve this error?

Sort rows in data.table in decreasing order on string key `order(-x,v)` gives error on data.table 1.9.4 or earlier

How do you refer to variables in a `data.table` if the variable names are stored in a character vector? For instance, this works for a `data.frame`:


    df &lt;- data.frame(col1 = 1:3)
    colname &lt;- &quot;col1&quot;
    df[colname] &lt;- 4:6
    df
    #   col1
    # 1    4
    # 2    5
    # 3    6

&lt;/pre&gt;

How can I perform this same operation for a data.table, either with or without `:=` notation? The obvious thing of `dt[ , list(colname)]` doesn&#39;t work (nor did I expect it to).   

Select / assign to data.table when variable names are stored in a character vector

I have always worked with the `zoo` package, that I have installed a long time ago. Today, I created a new R script, and ran `library(zoo)` and got the following error:

     &gt; library(zoo)
     Error in library(zoo) : there is no package called ‘zoo’

Odd.. Still, I tried reinstalling the package using `install.packages(&quot;zoo&quot;)`, and get the following:

     &gt; install.packages(&quot;zoo&quot;)
     Installing package(s) into ‘C:/Users/U122337.BOSTONADVISORS/Documents/R/win-library/2.15’
     (as ‘lib’ is unspecified)
     --- Please select a CRAN mirror for use in this session ---
     trying URL &#39;http://cran.cnr.Berkeley.edu/bin/windows/contrib/2.15/zoo_1.7-10.zip&#39;
     Content type &#39;application/zip&#39; length 874474 bytes (853 Kb)
     opened URL
     downloaded 853 Kb
  
     package ‘zoo’ successfully unpacked and MD5 sums checked
     Warning: cannot remove prior installation of package ‘zoo’

    The downloaded binary packages are in
        C:\Users\U122337.BOSTONADVISORS\AppData\Local\Temp\Rtmp404t8Y\downloaded_packages

When I try uploading the package, I get the error again that the `zoo` package doesn&#39;t exist.

I have no idea what&#39;s happening. I exited the GUI and restarted it, same problem. I have always worked with this package, and I have no idea why this is happening now.


Content Type	Original Author	Original Content on Stackoverflow
Question	geneorama	View Question on Stackoverflow
Solution 1 - R	Matt Dowle	View Answer on Stackoverflow
Solution 2 - R	TMo	View Answer on Stackoverflow

How to exclude one column from data.table OR convert to data.table to MTS

R Problem Overview

R Solutions

Solution 1 - R

Solution 2 - R

How to ignore parent element's overflow:hidden in css

Insert using LEFT JOIN and INNER JOIN

Attributions